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GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 5, 1999, 471-488
ON SOME ENTIRE MODULAR FORMS OF AN
INTEGRAL WEIGHT FOR THE CONGRUENCE
SUBGROUP Γ0 (4N )
G. LOMADZE
Abstract. Four types of entire modular forms of weight 2s are constructed for the congruence subgroup Γ0 (4N ) when s is even. One
can find these forms helpful in revealing the arithmetical meaning of
additional terms in the formulas for the number of representations of
positive integers by positive quadratic forms with integral coefficients
in an even number of variables.
All the notations, definitions and lemmas used in this paper are taken
from [1].
To make the notations shorter throughout the paper we put
Ψj (τ ; gl , hl , cl , Nl ) =
= Ψj τ ; g1 , h1 , c1 , N1 ; g2 , h2 , c2 , N2 ; . . . ; gs−8 , hs−8 , cs−8 , Ns−8
(j = 1, 2, 3, 4).
1.
Lemma 1. For even s ≥ 12 and given N let
1 ′′′
(τ ; c1 , 2N1 )ϑ′g2 h2 (τ ; c2 , 2N2 ) −
ϑ
N1 g 1 h 1
s−8
Y
1 ′′′
ϑg2 h2 (τ ; c2 , 2N2 )ϑ′g1 h1 (τ ; c1 , 2N1 )
ϑgk hk (τ ; ck , 2Nk ) (1.1)
−
N2
Ψ1 (τ ; gl , hl , cl , Nl ) =
k=3
1991 Mathematics Subject Classification. 11F11, 11F27.
Key words and phrases. Positive quadratic form, entire modular form, theta functions
with characteristics, congruence subgroup.
471
c 1999 Plenum Publishing Corporation
1072-947X/99/0900-0471$16.00/0
472
G. LOMADZE
and
Ψ2 (τ ; gl , hl , cl , Nl ) =
1 (4)
ϑ
(τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N12 g1 h1
1 (4)
(τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 ) −
ϑ
N22 g2 h2
s−8
Y
′′
′′
ϑg1 h1 (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 )
ϑgk hk (τ ; ck , 2Nk ), (1.2)
+
−
6
N1 N2
k=3
where
s−8
X
hk
.
2|gk , Nk |N (k = 1, 2, . . . , s − 8), 4N
Nk
(1.3)
k=1
Then we have
(γτ + δ)s/2 Ψj (τ ; gl , hl , 0, Nl ) =
n ατ + β
(j = 1, 2) (1.4)
Bn(j) e
4N γτ + δ
n=0
∞
X
for all substitutions from Γ in the neighborhood of each rational point τ =
6 0, (γ, δ) = 1).
− γδ (γ =
In [2] this lemma is proved for arbitrary s > 9 (even and odd).
Lemma 2. For even s we have
isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2 =
Proof. (a) Let s ≡ 0 (mod 4). Then
1
sgn δ
−1
|δ|
if s ≡ 0 (mod 4),
if s ≡ 2 (mod 4).
isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2 = i2hη(γ)(sgn δ−1) · i(2h−4)(1−|δ|) =
= (−1)hη(γ)(sgn δ−1) · (−1)(h−2)(1−|δ|) = 1.
(b) Let s ≡ 2 (mod 4). Then
isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2 = i(4h+2)η(γ)(sgn δ−1)/2 · i(4h−6)(1−|δ|)/2 =
−1
.
= i(2h+1)η(γ)(sgn δ−1) · (−1)(2h−3)(1−|δ|)/2 = sgn δ
|δ|
Theorem 1. For even s and given N the functions Ψj (τ ; gl , hl , 0, Nl )
(j = 1, 2) are entire modular forms of weight 2s and the character
∆
if s ≡ 0 (mod 4),
|δ|
(1.5)
−∆
χ(δ) =
sgn δ
if s ≡ 2 (mod 4)
|δ|
ON SOME ENTIRE MODULAR FORMS
473
(∆ is the determinant of the positive quadratic form with integral coefficients
in s variables for which the function Ψj will be used) for the congruence
subgroup Γ0 (4N ) if the following conditions hold:
1) 2|gk , Nk |N (k = 1, 2, . . . , s − 8),
s−8
X
s−8
gk2
X h2k
,
, 4
2) 4N
Nk
4Nk
3)
k=1
s−8
Q
Nk
k=1
|δ|
(1.6)
(1.7)
k=1
Ψj (τ ; αgl , hl , 0, Nl ) =
∆
|δ|
Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) (1.8)
for all α and δ with αδ ≡ 1 (mod 4N ).
Proof. I. As is well known, the theta-series which contained in the functions
Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) are regular on H. Therefore these functions
satisfy condition 1) and by Lemma 1 also condition 4) of the Definition
from [1] (p. 53).
II. From (1.6) it follows that
Γ0 (4N ) ⊂ Γ0 (4Nk ) (k = 1, 2, . . . , s − 8).
(1.9)
It is easy to verify that (1.7) implies
s−8
s−8 2
X
X
gk2 2ϕ(2Nk )−2
hk
, 4
δ
4N δ 2
,
Nk
4Nk
(1.10)
k=1
k=1
since 2 ∤ δ by virtue of αδ ≡ 1 (mod 4N ).
Using (1.9), (1.10) and Lemma 3 from [1] (p. 58), for all substitutions
from Γ0 (4N ) we get:
1) if n = 3, n = 1 and n = 0,
ατ + β
ατ + β
; 0, 2Nr ϑ′gt ht
; 0, 2Nt ×
ϑ′′′
gr hr
γτ + δ
γτ + δ
s−8
Y
ατ + β
ϑgk hk
; 0, 2Nk =
×
γτ + δ
k=3
=i
sη(γ)(sgn δ−1)/2
(s−8)(1−|δ|)/2
·i
s−8
Q
Nk
(γτ + δ)s/2 ×
|δ|
k=1
×ϑ′′′
αgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt )
s−8
Y
k=3
for r = 1, t = 2 and r = 2, t = 1;
ϑαgk ,hk (τ ; 0, 2Nk )
(1.11)
474
G. LOMADZE
2) if n = 4 and n = 0,
(4)
ϑgr hr
ατ + β
ατ + β
; 0, 2Nr ϑgt ht
; 0, 2Nt ×
γτ + δ
γτ + δ
s−8
Y
ατ + β
; 0, 2Nk =
×
ϑgk hk
γτ + δ
k=3
=i
sη(γ)(sgn δ−1)/2
(s−8)(1−|δ|)/2
·i
s−8
Q
Nk
(γτ + δ)s/2 ×
|δ|
k=1
(4)
×ϑαgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt )
s−8
Y
ϑαgk ,hk (τ ; 0, 2Nk )
(1.12)
k=3
for r = 1, t = 2 and r = 2, t = 1;
3) if n = 2 and n = 0,
ϑg′′1 h1
ατ + β
ατ + β
; 0, 2N1 ϑ′′g2 h2
; 0, 2N2 ×
γτ + δ
γτ + δ
s−8
ατ + β
Y
; 0, 2Nk =
ϑgk hk
×
γτ + δ
k=3
= isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2
s−8
Q
Nk
(γτ + δ)s/2 ×
|δ|
k=1
×ϑ′′αg1 ,h1 (τ ; 0, 2N1 )ϑ′′αg2 ,h2 (τ ; 0, 2N2 )
s−8
Y
ϑαgk ,hk (τ ; 0, 2Nk ).
(1.13)
k=3
Thus by (1.1), (1.11) and (1.2), (1.12), (1.13) we have
Ψj
ατ + β
; gl , hl , 0, Nl = isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2 ×
γτ + δ
s−8
Q
Nk
× k=1
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) (j = 1, 2),
|δ|
which by Lemma 2 implies respectively for j = 1 and j = 2
Ψj
ατ + β
; gl , hl , 0, Nl =
γτ + δ
s−8
Q
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl )
|δ|
if s ≡ 0 (mod 4)
k=1
ON SOME ENTIRE MODULAR FORMS
475
and
Ψj
= sgn δ
−1
|δ|
ατ + β
; gl , hl , 0, Nl =
γτ + δ
s−8
Q
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 2 (mod 4).
|δ|
k=1
Hence by (1.5) and (1.8) we obtain
ατ + β
Ψj
; gl , hl , 0, Nl = χ(δ)(γτ + δ)s/2 Ψj (τ ; gl , hl , 0, Nl ).
γτ + δ
Thus the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) satisfy condition (2) of the
Definition from [1].
III. The functions Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) also satisfy condition 3)
of the Definition from [1]. For the proof see [2] (p. 109).
2.
To simplify the notations of this section we put
Ω(τ ; g1 , h1 , c1 , N1 ; g2 , h2 , h2 , c2 , N2 ) =
1 ′′
=
(τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) −
ϑ
N1 g1 h1
1 ′′
−
ϑ
(τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 )
N2 g2 h2
(2.1)
and occasionally
(s − 2)/3 = se, (s − 1)/3 = s′ .
(2.2)
Lemma 3. For even s ≥ 10 and given N let
Ψ3 (τ ; gl , hl , cl , Nl ) = Ω(τ ; g1 , h1 , c1 , N1 ; g2 , h2 , c2 , N2 ) ×
×
s/3
Y
ϑ′gk hk (τ ; ck , 2Nk ) if s ≡ 0 (mod 6),
(2.3)
k=3
Ψ3 (τ ; gl , hl , cl , Nl ) = Ω(τ ; g1 , h1 , c1 , N1 ; g2 , h2 , c2 , N2 ) ×
×
s˜
Y
ϑg′ k hk (τ ; ck , 2Nk )
k=3
+2
s˜
Y
ϑgk hk (τ ; ck , 2Nk ) if s ≡ 2 (mod 6),
(2.4)
s+1
k=˜
Ψ3 (τ ; gl , hl , cl , Nl ) = Ω(τ ; g1 , h1 , c1 , N1 ; g2 , h2 , c2 , N2 ) ×
′
×
s
Y
k=3
ϑ′gk hk (τ ; ck , 2Nk )ϑgs′ +1 hs′ +1 (τ ; cs′ +1 , 2Ns′ +1 ) if s ≡ 4 (mod 6) (2.5)
476
G. LOMADZE
and
Ψ4 (τ ; gl , hl , cl , Nl ) =
s/3
Y
ϑ′gk hk (τ ; ck , 2Nk ) if s ≡ 0 (mod 6),
(2.6)
k=1
Ψ4 (τ ; gl , hl , cl , Nl ) =
s˜
Y
ϑ′gk hk (τ ; ck , 2Nk )ϑgs˜+1 hs˜+1 (τ ; cs˜+1 , 2Ns˜+1 ) ×
k=1
×ϑgs˜+2 hs˜+2 (τ ; cs˜+2 , 2Ns˜+2 ) if s ≡ 2 (mod 6),
(2.7)
′
Ψ4 (τ ; gl , hl , cl , Nl ) =
s
Y
ϑ′gk hk (τ ; ck , 2Nk )ϑgs′ +1 hs′ +1 (τ ; cs′ +1 , 2Ns′ +1 ) (2.8)
k=3
if s ≡ 4 (mod 6),
where
s/3
X hk
4N
if s ≡ 0 (mod 6),
Nk
2|gk , Nk |N (k = 1, 2, . . . , s/3),
(2.9)
k=1
2|gk , Nk |N (k = 1, 2, . . . , se + 2),
2|gk , Nk |N (k = 1, 2, . . . , s′ + 1),
s˜+2
X
hk
if s ≡ 2 (mod 6), (2.10)
4N
Nk
k=1
′
+1
sX
hk
if s ≡ 4 (mod 6), (2.11)
4N
Nk
k=1
Then we have
(γτ + δ)s/2 Ψj (τ ; gl , hl , 0, Nl ) =
n ατ + β
(j = 3, 4)
Cn(j) e
4N γτ + δ
n=0
∞
X
(2.12)
for all substitutions from Γ in the neighborhood of each point τ = − γδ (γ 6= 0,
(γ, δ) = 1).
Proof. I. It is shown in [3] (formula (1.7)) that
(γτ + δ)3 Ω(τ ; g1 , h1 , 0, N1 ; g2 , h2 , 0, N2 ) =
3
= −e
sgn γ (4N1 N2 γ 2 )−1/2
4
×Ω
X
2
Y
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
H1 mod 2N1 k=1
H2 mod 2N2
ατ + β
; g1′ h1′ , H1 , N1 ; g2′ , h′2 , H2 , 2N2 .
γτ + δ
For n = 1 and n = 0 it follows by Lemma 4 from [1] (p. 61) that:
(2.13)
ON SOME ENTIRE MODULAR FORMS
(γτ + δ)(s−6)/2
1)
s/3
Y
477
ϑg′ k hk (τ ; 0, 2Nk ) =
k=3
× 2(s−6)/3
= e (s − 6)(sgn γ)/8 (−i sgn γ)(s−6)/3 ×
s/3
s/3
−1/2
Y
Y
X
(s−6)/3
Nk |γ|
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
k=3
H3 mod 2N3 k=3
........................
Hs/3 mod 2Ns/3
ατ + β
if s ≡ 0 (mod 6),
; Hk , 2Nk
×ϑgk′ h′k
γτ + δ
2)
(γτ + δ)(s−6)/2
s˜
Y
(2.14)
ϑ′gk hk (τ ; 0, 2Nk )ϑgs˜+1 hs˜+1 (τ ; 0, 2Ns˜+1 )×
k=3
×ϑgs˜+2 hs˜+2 (τ ; 0, 2Ns˜+2 ) = e (s − 6)(sgn γ)/8 ×
s˜+2
−1/2
X
× 2s˜
(−i sgn γ)(s−8)/3 ×
Nk |γ|s˜
k=3
s˜
Y
X
×
k
k=3
H3 mod 2N3
........................
Hs˜+2 mod 2Ns˜+2
×
s˜
+2
Y
k=˜
s+1
ατ + β
×
; Hk , 2Nk
k
γτ + δ
ϕgk′ gk hk (0, Hk ; 2Nk )ϑ′g′ h′
ϕgk′ gk hk (0, Hk ; 2Nk )ϑgk′ h′k
ατ + β
; Hk , 2Nk
γτ + δ
(2.15)
if s ≡ 2 (mod 6),
′
3)
(s−6)/2
(γτ + δ)
s
Y
ϑ′gk hk (τ ; 0, 2Nk )ϑgs′ +1 hs′ +1 (τ ; 0, 2Ns′ +1 ) =
k=3
+1
−1/2
′ sX
′
= e (s − 6)(sgn γ)/8 2s −1
Nk |γ|s −1
(−i sgn γ)(s−7)/3 ×
′
k=3
×
X
s′
Y
k=3
H3 mod 2N3
...........................
Hs′ +1 mod 2Ns′ +1
ατ + β
; Hk , 2Nk ×
k
γτ + δ
ϕgk′ gk hk (0, Hk ; 2Nk )ϑg′ ′ h′
k
×ϕg′ ′ gs′ +1 hs′ +1 (0, Hs′ +1 ; 2Ns′ +1 ) ×
s +1
ατ + β
×ϑgs′ +1 hs′ +1
; Hs′ +1 , 2Ns′ +1 if s ≡ 4 (mod 6).
γτ + δ
(2.16)
478
G. LOMADZE
After multiplying (2.13) successively by (2.14), (2.15) and (2.16), by (2.3)
we obtain
1) (γτ + δ)s/2 Ψ3 (τ ; gl , hl , 0, Nl ) = (−1)s/6 e s(sgn γ)/8 ×
s/3
−1/2
Y
s/3
Nk |γ|s/3
× 2
k=1
s/3
Y
X
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
H1 mod 2N1 k=3
........................
Hs/3 mod 2Ns/3
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 0 (mod 6),
γτ + δ
(γτ + δ)s/2 Ψ3 (τ ; gl , hl , 0, Nl ) = (−1)(s−2)/6 e s(sgn γ)/8 ×
×Ψ3
2)
s˜
+2
−1/2
Y
Nk |γ|s˜+2
× 2s˜+2
k=1
s˜
+2
Y
X
(2.17)
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
k=3
H1 mod 2N1
........................
Hs˜+2 mod 2Ns˜+2
ατ + β
(2.18)
; gl′ , h′l , Hl , Nl if s ≡ 2 (mod 6),
γτ + δ
(γτ + δ)s/2 Ψ3 (τ ; gl , hl , 0, Nl ) = (−1)(s+2)/6 (i sgn γ)e s(sgn γ)/8 ×
×Ψ3
3)
s′ +1
× 2
′
+1
sY
Nk |γ|
s′ +1
k=1
×Ψ3
−1/2
′
sY
+1
X
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
H1 mod 2N1
k=3
...........................
Hs′ +1 mod 2Ns′ +1
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 4 (mod 6).
γτ + δ
(2.19)
Further, reasoning as in [1] (Lemma 5, the bottom part of p. 64 and p.
65), we obtain (2.12) if j = 3.
II. By Lemma 4 from [1], for n = 1 and n = 0 it respectively follows by
(2.6), (2.7) and (2.8) that
s/2
(γτ + δ)
s/2
Ψ4 (τ ; gl , hl , 0, Nl ) = (γτ + δ)
s/3
Y
ϑ′gk hk (τ ; 0, 2Nk ) =
k=1
= e s(sgn γ)/8
×
X
s/3
2s/3
Y
k=1
Nk |γ|s/3
−1/2
(−i sgn γ)s/3 ×
ατ + β
Y
′
=
ϕgk′ gk hk (0, Hk ; 2Nk )ϑg′ h′
; Hk , 2Nk
k k
γτ + δ
s/3
H1 mod 2N1 k=1
........................
Hs/3 mod 2Ns/3
ON SOME ENTIRE MODULAR FORMS
479
s/3
−1/2
Y
×
= (−1)s/6 e s(sgn γ)/8 2s/3
Nk |γ|s/3
k=1
X
×
s/3
Y
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
H1 mod 2N1 k=1
........................
Hs/3 mod 2Ns/3
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 0 (mod 6),
(2.20)
γτ + δ
s˜
Y
(γτ + δ)s/2 Ψ4 (τ ; gl , hl , 0, Nl ) = (γτ + δ)s/2
ϑg′ k hk (τ ; 0, 2Nk ) ×
×Ψ4
k=1
×ϑgs˜+1 hs˜+1 (τ ; 0, 2Ns˜+1 )ϑgs˜+2 hs˜+2 (τ ; 0, 2Ns˜+2 ) =
s˜
+2
−1/2
Y
(−i sgn γ)s˜ ×
Nk |γ|s˜+2
= e s(sgn γ)/8 2s˜+2
k=1
×
s˜
Y
X
H1 mod 2N1 k=1
..................
Hs˜ mod 2Ns˜
×
ατ + β
×
; Hk , 2Nk
ϕgk′ gk hk (0, Hk ; 2Nk )ϑ′g′ h′
k k
γτ + δ
+2
s˜
Y
X
s+1
Hs˜+1 mod 2Ns˜+1 k=˜
Hs˜+2 mod 2Ns˜+2
ατ + β
=
; Hk , 2Nk
ϕgk′ gk hk (0, Hk ; 2Nk )ϑgk hk
γτ + δ
s˜
+2
−1/2
Y
= (−1)(s−2)/6 e s(sgn γ)/8 2s˜+2
×
Nk |γ|s˜+2
k=1
×
X
s˜
+2
Y
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
k=1
H1 mod 2N1
........................
Hs˜+2 mod 2Ns˜+2
×Ψ4
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 2 (mod 6)
γτ + δ
(2.21)
and
′
(γτ + δ)
s/2
Ψ4 (τ ; gl , hl , 0, Nl ) = (γτ + δ)
s/2
s
Y
ϑg′ k hk (τ ; 0, 2Nk ) ×
k=1
×ϑgs′ +1 hs′ +1 (τ ; 0, 2Ns′ +1 )e s(sgn γ)/8 ×
′
+1
−1/2
′ sY
′
′
s +1
(−i sgn γ)s ×
Nk |γ|s +1
× 2
k=1
480
×
G. LOMADZE
s′
ατ + β
Y
′
×
; Hk , 2Nk
ϕgk′ gk hk (0, Hk ; 2Nk )ϑg′ h′
k k
γτ + δ
X
H1 mod 2N1 k=1
.....................
Hs′ mod 2Ns′
X
×
′
ϕgs+1
gs′ +1 hs′ +1 (0, Hs′ +1 ; 2Ns′ +1 ) ×
Hs′ +1 mod 2Ns′ +1
ατ + β
; Hs′ +1 , 2Ns′ +1 =
γτ + δ
′
+1
−1/2
s′ +1 sY
′
(s+2)/6
(i sgn γ)e s(sgn γ)/8 2
= (−1)
Nk |γ|s +1
×
×ϑgs′ +1 hs′ +1
k=1
×
X
′
sY
+1
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
k=1
H1 mod 2N1
.....................
Hs′ +1 mod 2Ns′ +1
×Ψ4
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 4 (mod 6).
γτ + δ
(2.22)
Further, reasoning as in [1] (Lemma 5), we obtain (2.12) if j = 4.
Lemma 4. For even s we have
(a) isη(γ)(sgn δ−1)/2 · i(s/3)(1−|δ|)/2 =
1
−1
if s ≡ 0 (mod 12),
sgn δ
if s ≡ 6 (mod 12);
|δ|
1
if s ≡ 8 (mod 12),
−1
(b) isη(γ)(sgn δ−1)/2 · i((s+4)/3)(1−|δ|)/2=
sgn δ
if s ≡ 2 (mod 12);
|δ|
−1
if s ≡ 4 (mod 12),
sη(γ)(sgn δ−1)/2
((s+2)/3)(1−|δ|)/2
|δ|
·i
(c) i
=
sgn δ
if s ≡ 10 (mod 12).
Proof. (a) Let s ≡ 0 (mod 6), i.e., s ≡ 0 (mod 12) or s ≡ 6 (mod 12).
Then
isη(γ)(sgn δ−1)/2 · i(s/3)(1−|δ|)/2 = i3hη(γ)(sgn δ−1) · i2h(1−|δ|)/2 =
1
if 2|h, i.e., if s ≡ 0 (mod 12),
−1
=
sgn δ
if 2 ∤ h, i.e., if s ≡ 6 (mod 12).
|δ|
ON SOME ENTIRE MODULAR FORMS
481
(b) Let s ≡ 2 (mod 6), i.e., s ≡ 2 (mod 12) or s ≡ 8 (mod 12). Then
isη(γ)(sgn δ−1)/2 · i((s+4)/3)(1−|δ|)/2 = i(3h+1)η(γ)(sgn δ−1) · i2(h+1)(1−|δ|)/2 =
= i(h+1)η(γ)(sgn δ−1) · (−1)(h+1)(1−|δ|)/2 =
sgn δ −1
if 2|h, i.e., if s ≡ 2 (mod 12),
|δ|
=
1
if 2 ∤ h, i.e., if s ≡ 8 (mod 12).
(c) Let s ≡ 4 (mod 6), i.e., s ≡ 4 (mod 12) or s ≡ 10 (mod 12). Then
isη(γ)(sgn δ−1)/2 · i((s+2)/3)(1−|δ|)/2 = i(3h+2)η(γ)(sgn δ−1) · i2(h+1)(1−|δ|)/2 =
−1
if 2|h, i.e., if s ≡ 4 (mod 12),
|δ|
=
if 2 ∤ h, i.e., if s ≡ 10 (mod 12).
sgn δ
Theorem 2. For even s and given N the functions Ψj (τ ; gl , hl , 0, Nl )
(j = 3, 4) are entire modular forms of weight 2s and the character
∆
if s ≡ 0 (mod 4),
|δ|
−∆
(2.23)
χ(δ) =
sgn δ
if s ≡ 2 (mod 4)
|δ|
(∆ is the determinant of the positive quadratic form with integral coefficients
in s variables for which the function Ψj will be used) for the congruence
subgroup Γ0 (4N ) if the following conditions hold:
(a) when s ≡ 0 (mod 6)
1) 2|gk , Nk |N (k = 1, 2, . . . , s/3),
s/3
X
hk2
,
2) 4N
Nk
k=1
(2.24)
s/3
X
gk2
4
,
4Nk
(2.25)
k=1
3) for all α and δ with αδ ≡ 1 (mod 4N )
s/3
Q
k=1
Nk
|δ|
Ψj (τ ; αgl , hl , 0, Nl ) =
∆
|δ|
Ψj (τ ; gl , hl , 0, Nl );
(2.26)
(b) when s ≡ 2 (mod 6)
1) 2|gk , Nk |N (k = 1, 2, . . . , (s + 4)/3),
(s+4)/3
(s+4)/3
X h2
X gk2
k
, 4
,
2) 4N
Nk
4Nk
k=1
k=1
(2.27)
(2.28)
482
G. LOMADZE
3) for all α and δ with αδ ≡ 1 (mod 4N )
(s+4)/3
Q
Nk
∆
Ψj (τ ; αgl , hl , 0, Nl ) =
Ψj (τ ; gl , hl , 0, Nl );
|δ|
|δ|
k=1
(2.29)
(c) when s ≡ 4 (mod 6)
1) 2|gk , Nk |N (k = 1, 2, . . . , (s + 2)/3),
(s+2)/3
X h2
k
2) 4N
,
Nk
k=1
(s+2)/3
X gk2
4
,
4Nk
(2.30)
(2.31)
k=1
3) for all α and δ with αδ ≡ 1 (mod 4N )
(s+2)/3
Q
k=1
|δ|
Nk
Ψj (τ ; αgl , hl , 0, Nl ) = sgn δ
−∆
Ψj (τ ; gl , hl , 0, Nl ). (2.32)
|δ|
Proof. I. As in the case of Theorem 1, the functions Ψj (τ ; gl , hl , 0, Nl ) (j =
3, 4) satisfy condition 1) and by Lemma 3 also condition 4) of the Definition
from [1].
II. From (2.24), (2.27) and (2.30) we obtain
Γ0 (4N ) ⊂ Γ0 (4Nk ).
(2.33)
Obviously, (2.25), (2.28) and (2.31) respectively imply
s/3
2 X h2k
,
4N δ
Nk
k=1
(s+4)/3 2
X h
k
4 N δ 2
,
Nk
k=1
and
(s+2)/3 2
X h
k
,
4N δ 2
Nk
k=1
s/3
X
gk2 2ϕ(2Nk )−2
δ
if s ≡ 0 (mod 6),
4
4Nk
(2.34)
k=1
(s+4)/3
X gk2 2ϕ(2Nk )−2
δ
4
if s ≡ 2 (mod 6), (2.35)
4Nk
k=1
(s+2)/3
X gk2 2ϕ(2Nk )−2
δ
4
if s ≡ 4 (mod 6).
4Nk
(2.36)
k=1
Using (2.33) and Lemma 3 from [1], for all substitutions from Γ0 (4N ) we
obtain:
1) for n = 2, n = 1 and n = 0
ατ + β
ατ + β
ϑ′′gr hr
; 0, 2Nr ϑgt ht
; 0, 2Nt ×
γτ + δ
γτ + δ
ON SOME ENTIRE MODULAR FORMS
×
s/3
Y
k=3
ϑ′gk hk
483
ατ + β
; 0, 2Nk =
γτ + δ
s/3
Q
Nk
(γτ + δ)s/2 ×
·i
|δ|
×ϑ′′αgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt ) ×
sη(γ)(sgn δ−1)/2
=i
×
s/3
Y
(s/3)(1−|δ|)/2
k=1
′
(τ ; 0, 2Nk ) if s ≡ 0 (mod 6),
ϑαg
k ,hk
(2.37)
k=3
ατ + β
ατ + β
; 0, 2Nr ϑgt ht
; 0, 2Nt ×
γτ + δ
γτ + δ
s
˜
ατ + β
Y
ϑ′gk hk
×
; 0, 2Nk ×
γτ + δ
k=3
ατ + β
ατ + β
; 0, 2Ns˜+1 ϑgs˜+2 hs˜+2
; 0, 2Ns˜+2 =
×ϑgs˜+1 hs˜+1
γτ + δ
γτ + δ
s˜Q
+2
Nk
= isη(γ)(sgn δ−1)/2 · i(˜s+2)(1−|δ|)/2 k=1
(γτ + δ)s/2 ×
|δ|
ϑ′′gr hr
′′
×ϑαg
(τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt )
r ,hr
s˜
Y
ϑ′αgk ,hk (τ ; 0, 2Nk ) ×
k=3
×ϑαgs˜+1 ,hs˜+1 (τ ; 0, 2Ns˜+1 )ϑαgs˜+2 ,hs˜+2 (τ ; 0, 2Ns˜+2 )
(2.38)
if s ≡ 2 (mod 6),
ατ + β
ατ + β
ϑg′′r hr
; 0, 2Nr ϑgt ht
; 0, 2Nt ×
γτ + δ
γτ + δ
′
s
ατ + β
Y
ατ + β
; 0, 2Nk ϑgs′ +1 hs′ +1
; 0, 2Ns′ +1 =
×
ϑg′ k hk
γτ + δ
γτ + δ
k=3
′
= (sgn δ)isη(γ)(sgn δ−1)/2 · i(s +1)(1−|δ|)/2 ×
′
s
Q
Nk
k=1
×
(γτ + δ)s/2 ϑ′′αgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt ) ×
|δ|
′
×
s
Y
ϑαgk ,hk (τ ; 0, 2Nk )ϑαgs′ +1 ,hs′ +1 (τ ; 0, 2Ns′ +1 )
k=3
if s ≡ 4 (mod 6),
(2.39)
484
G. LOMADZE
where r = 2, t = 1 and r = 1, t = 2;
2) for n = 1 and n = 0
s/3
Y
ϑ′gk hk
k=1
ατ + β
, 0, 2Nk =
γτ + δ
= isη(γ)(sgn δ−1)/2 · i(s/3)(1−|δ|)/2
s/2
×(γτ + δ)
s/3
Y
s/3
Q
k=1
Nk
|δ|
×
ϑ′αgk ,hk (τ ; 0, 2Nk ) if s ≡ 0 (mod 6),
(2.40)
k=1
s˜
Y
ϑ′gk hk
k=1
ατ + β
ατ + β
, 0, 2Nk ϑgs˜+1 hs˜+1
, 0, 2Ns˜+1 ×
γτ + δ
γτ + δ
ατ + β
, 0, 2Ns˜+2 = isη(γ)(sgn δ−1)/2 · i(˜s+2)(1−|δ|)/2 ×
γτ + δ
s˜Q
+2
Nk
s˜
Y
k=1
×
ϑ′αgk ,hk (τ ; 0, 2Nk ) ×
(γτ + δ)s/2
|δ|
×ϑgs˜+2 hs˜+2
k=1
×ϑαgs˜+1 ,hs˜+1 (τ ; 0, 2Ns˜+1 )ϑαgs˜+2 ,hs˜+2 (τ ; 0, 2Ns˜+2 )
(2.41)
if s ≡ 2 (mod 6),
s′
Y
ϑ′gk hk
k=1
ατ + β
ατ + β
, 0, 2Nk ϑgs′ +1 hs′ +1
, 0, 2Ns′ +1 =
γτ + δ
γτ + δ
′
= sgn δisη(γ)(sgn δ−1)/2 · i(s +1)(1−|δ|)/2
′
sQ
+1
k=1
Nk
|δ|
(γτ + δ)s/2 ×
′
×
s
Y
ϑ′αgk hk (τ ; 0, 2Nk )ϑαgs′ +1 ,hs′ +1 (τ ; 0, 2Ns′ +1 )
k=1
if s ≡ 4 (mod 6).
By (2.3), (2.37) and (2.6), (2.40), for j = 3 and j = 4 we get
Ψj
ατ + β
; gl , hl , 0, Nl = isη(γ)(sgn δ−1)/2 · i(s/3)(1−|δ|)/2 ×
γτ + δ
(2.42)
ON SOME ENTIRE MODULAR FORMS
×
485
s/3
Q
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 0 (mod 6). (2.43)
|δ|
k=1
By (2.4), (2.38) and (2.7), (2.41), for j = 3 and j = 4 we get
Ψj
×
ατ + β
; gl , hl , 0, Nl = isη(γ)(sgn δ−1)/2 · i((s+4)/3)(1−|δ|)/2 ×
γτ + δ
(s+4)/3
Q
k=1
|δ|
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 2 (mod 6). (2.44)
By (2.5), (2.39) and (2.8), (2.42), for j = 3 and j = 4 we get
Ψj
×
ατ + β
; gl , hl , 0, Nl = isη(γ)(sgn δ−1)/2 · i((s+2)/3)(1−|δ|)/2 ×
γτ + δ
(s+2)/3
Q
k=1
|δ|
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 4 (mod 6). (2.45)
Further, by Lemma 4, for j = 3, 4 realtions (2.43), (2.44), and (2.45)
respectively imply:
s/3
Q
Nk
ατ + β
k=1
Ψj
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl )
; gl , hl , 0, Nl =
γτ + δ
|δ|
if s ≡ 0 (mod 12),
−1
s/3
Q
k=1
(2.46)
Nk
(γτ + δ)s/2 ×
|δ|
|δ|
× Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 6 (mod 12);
= sgn δ
Ψj
(s+4)/3
Q
Nk
−1
ατ + β
k=1
(γτ + δ)s/2 ×
; gl , hl , 0, Nl = sgn δ
γτ + δ
|δ|
|δ|
× Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 2 (mod 12),
=
(s+4)/3
Q
k=1
|δ|
(2.47)
Nk
s/2
(γτ + δ) Ψj (τ ; αgl , hl , 0, Nl )
if s ≡ 8 (mod 12);
486
G. LOMADZE
(s+2)/3
Q
Nk
ατ + β
−1
k=1
(γτ + δ)s/2 ×
; gl , hl , 0, Nl = sgn δ
Ψj
γτ + δ
|δ|
|δ|
× Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 4 (mod 12),
=
(s+2)/3
Q
k=1
|δ|
(2.48)
Nk
s/2
(γτ + δ) Ψj (τ ; αgl , hl , 0, Nl )
if s ≡ 10 (mod 12).
From (2.46)–(2.48), (2.26), (2.29), (2.32), and (2.23) it follows that
Ψj
ατ + β
; gl , hl , 0, Nl = χ(δ)(γτ + δ)s/2 Ψj (τ ; gl , hl , 0, Nl ) (j = 3, 4).
γτ + δ
Thus the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 3, 4) satisfy condition 2) of the
Definition from [1].
III. By [1] (formula 1.3), for r = 1, t = 2 and r = 2, t = 1 we obtain:
1) If s ≡ 0 (mod 6),
ϑ′′gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )
s/3
Y
ϑ′gk hk (τ ; 0, 2Nk ) =
k=3
= (πi)s/3
∞
X
hr mr +ht mt +
(−1)
s/3
P
k=3
hk mk
(4Nr mr + gr )2 ×
mr ,mt ,m3 ,...,ms/3 =−∞
×
s/3
Y
k=3
X
X
s/3
∞
gk 2
1
=
2Nk mk +
Bn(r,t) e(nτ ),
(4Nk mk + gk )e
4Nk
2
n=0
k=1
since by (2.24) and (2.25),
n=
s/3
s/3
s/3
X
1X 2
1
gk 2 X
Nk m2k + mk gk /2 +
gk /4Nk
=
2Nk mk +
4Nk
2
4
k=1
k=1
k=1
is a non-negative integer;
2) If s ≡ 2 (mod 6),
ϑg′′r hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )
s˜
Y
ϑ′gk hk (τ ; 0, 2Nk ) ×
k=3
×ϑgs˜+1 hs˜+1 (τ ; 0, 2Ns˜+1 )ϑgs˜+2 hs˜+2 (τ ; 0, 2Ns˜+2 ) =
ON SOME ENTIRE MODULAR FORMS
P
487
˜+2
s
∞
X
s˜
= (πi)
hr mr +ht mt +
(−1)
hk mk
k=3
(4Nr mr + gr )2 ×
mr ,mt ,m3 ,...,ms˜+2 =−∞
s˜
Y
×
k=3
X
X
s˜+2
∞
1
gk 2
(4Nk mk + gk )e
=
Cn(r,t) e(nτ ),
2Nk mk +
4Nk
2
n=0
k=1
since by (2.27) and (2.28),
n=
s˜+2
s˜+2
s˜+2
X
1X
gk 2 X
1
gk2 /4Nk
2Nk mk +
Nk m2k + mk gk /2 +
=
4Nk
2
4
k=1
k=1
k=1
is a non-negative integer;
3) If s ≡ 4 (mod 6),
ϑ′′gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt ) ×
′
×
s
Y
ϑ′gk hk (τ ; 0, 2Nk )ϑgs′ +1 hs′ +1 (τ ; 0, 2Ns′ +1 ) ×
k=3
P
s′ +1
= (πi)
∞
X
′
s
hr mr +ht mt +
(−1)
k=3
h k mk
(4Nr mr + gr )2 ×
mr ,mt ,m3 ,...,ms′ +1 =−∞
′
×
s
Y
k=3
′
X
sX
∞
+1
gk 2
1
Dn(r,t) e(nτ ),
=
2Nk mk +
(4Nk mk + gk )e
4Nk
2
n=0
k=1
since By (2.30) and (2.31)
n=
′
sX
+1
k=1
′
+2
s′ +1
1 sX
gk 2 X
1
2
2Nk mk +
gk2 /4Nk
=
Nk mk + mk gk /2 +
4Nk
2
4
k=1
k=1
is a non-negative integer.
Analogously,
s/3
Y
ϑg′ k hk (τ ; 0, 2Nk ) =
k=1
s/3
= (πi)
∞
X
m1 ,m2 ,...,ms/3 =−∞
(−1)
s/3
P
k=1
s/3
hk mk Y
(4Nk mk + gk ) ×
k=1
s/3
∞
X
2 X
1
×e
Bn e(nτ ) if s ≡ 0 (mod 6),
2Nk mk + gk /2 τ =
4Nk
n=0
k=1
488
G. LOMADZE
s˜
Y
ϑ′gk hk (τ ; 0, 2Nk )ϑgs˜+1 hs˜+1 (τ ; 0, 2Ns˜+1 )ϑgs˜+2 hs˜+2 (τ ; 0, 2Ns˜+2 ) =
k=1
P
˜+2
s
∞
X
= (πi)s˜
hk mk
(−1)k=1
m1 ,m2 ,...,ms˜+2 =−∞
s˜
Y
(4Nk mk + gk ) ×
k=1
s˜+2
∞
X
2 X
1
×e
2Nk mk + gk /2 τ =
Cn e(nτ ) if s ≡ 2 (mod 6),
4Nk
n=0
k=1
′
s
Y
ϑg′ k hk (τ ; 0, 2Nk )ϑgs′ +1 hs′ +1 (τ ; 0, 2Ns′ +1 ) =
k=1
P
s′ +1
s
= (πi)
′
∞
X
(−1)
k=1
m1 ,m2 ,...,ms′ +1 =−∞
s′
hk mk Y
(4Nk mk + gk ) ×
k=1
′
∞
+1
sX
2 X
1
2Nk mk + gk /2 τ =
Dn e(nτ ) if s ≡ 4 (mod 6).
×e
4Nk
n=0
k=1
Thus by (2.3)–(2.8) the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 3, 4) satisfy
condition 3) of the Definition from [1].
References
1. G. Lomadze, On some entire modular forms of weights 5 and 6 for the
congruence group Γ0 (4N ). Georgian Math. J. 1(1994), No. 1, 53–76.
2. G. Lomadze, On some entire modular forms of half-integral weight for
the congruence group Γ0 (4N ). Proc. A. Razmadze Math. Inst. 118(1998),
101–109.
3. G. Lomadze, Construction of entire modular forms of weights 5 and
6 for the congruence group Γ0 (4N ). Georgian Math. J. 2(1995), No. 2,
189–199.
(Received 14.04.1998)
Author’s address:
Faculty of Mechanics and Mathematics
I. Javakhishvili Tbilisi State University
2, University St., Tbilisi 380043
Georgia
ON SOME ENTIRE MODULAR FORMS OF AN
INTEGRAL WEIGHT FOR THE CONGRUENCE
SUBGROUP Γ0 (4N )
G. LOMADZE
Abstract. Four types of entire modular forms of weight 2s are constructed for the congruence subgroup Γ0 (4N ) when s is even. One
can find these forms helpful in revealing the arithmetical meaning of
additional terms in the formulas for the number of representations of
positive integers by positive quadratic forms with integral coefficients
in an even number of variables.
All the notations, definitions and lemmas used in this paper are taken
from [1].
To make the notations shorter throughout the paper we put
Ψj (τ ; gl , hl , cl , Nl ) =
= Ψj τ ; g1 , h1 , c1 , N1 ; g2 , h2 , c2 , N2 ; . . . ; gs−8 , hs−8 , cs−8 , Ns−8
(j = 1, 2, 3, 4).
1.
Lemma 1. For even s ≥ 12 and given N let
1 ′′′
(τ ; c1 , 2N1 )ϑ′g2 h2 (τ ; c2 , 2N2 ) −
ϑ
N1 g 1 h 1
s−8
Y
1 ′′′
ϑg2 h2 (τ ; c2 , 2N2 )ϑ′g1 h1 (τ ; c1 , 2N1 )
ϑgk hk (τ ; ck , 2Nk ) (1.1)
−
N2
Ψ1 (τ ; gl , hl , cl , Nl ) =
k=3
1991 Mathematics Subject Classification. 11F11, 11F27.
Key words and phrases. Positive quadratic form, entire modular form, theta functions
with characteristics, congruence subgroup.
471
c 1999 Plenum Publishing Corporation
1072-947X/99/0900-0471$16.00/0
472
G. LOMADZE
and
Ψ2 (τ ; gl , hl , cl , Nl ) =
1 (4)
ϑ
(τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) +
N12 g1 h1
1 (4)
(τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 ) −
ϑ
N22 g2 h2
s−8
Y
′′
′′
ϑg1 h1 (τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 )
ϑgk hk (τ ; ck , 2Nk ), (1.2)
+
−
6
N1 N2
k=3
where
s−8
X
hk
.
2|gk , Nk |N (k = 1, 2, . . . , s − 8), 4N
Nk
(1.3)
k=1
Then we have
(γτ + δ)s/2 Ψj (τ ; gl , hl , 0, Nl ) =
n ατ + β
(j = 1, 2) (1.4)
Bn(j) e
4N γτ + δ
n=0
∞
X
for all substitutions from Γ in the neighborhood of each rational point τ =
6 0, (γ, δ) = 1).
− γδ (γ =
In [2] this lemma is proved for arbitrary s > 9 (even and odd).
Lemma 2. For even s we have
isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2 =
Proof. (a) Let s ≡ 0 (mod 4). Then
1
sgn δ
−1
|δ|
if s ≡ 0 (mod 4),
if s ≡ 2 (mod 4).
isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2 = i2hη(γ)(sgn δ−1) · i(2h−4)(1−|δ|) =
= (−1)hη(γ)(sgn δ−1) · (−1)(h−2)(1−|δ|) = 1.
(b) Let s ≡ 2 (mod 4). Then
isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2 = i(4h+2)η(γ)(sgn δ−1)/2 · i(4h−6)(1−|δ|)/2 =
−1
.
= i(2h+1)η(γ)(sgn δ−1) · (−1)(2h−3)(1−|δ|)/2 = sgn δ
|δ|
Theorem 1. For even s and given N the functions Ψj (τ ; gl , hl , 0, Nl )
(j = 1, 2) are entire modular forms of weight 2s and the character
∆
if s ≡ 0 (mod 4),
|δ|
(1.5)
−∆
χ(δ) =
sgn δ
if s ≡ 2 (mod 4)
|δ|
ON SOME ENTIRE MODULAR FORMS
473
(∆ is the determinant of the positive quadratic form with integral coefficients
in s variables for which the function Ψj will be used) for the congruence
subgroup Γ0 (4N ) if the following conditions hold:
1) 2|gk , Nk |N (k = 1, 2, . . . , s − 8),
s−8
X
s−8
gk2
X h2k
,
, 4
2) 4N
Nk
4Nk
3)
k=1
s−8
Q
Nk
k=1
|δ|
(1.6)
(1.7)
k=1
Ψj (τ ; αgl , hl , 0, Nl ) =
∆
|δ|
Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) (1.8)
for all α and δ with αδ ≡ 1 (mod 4N ).
Proof. I. As is well known, the theta-series which contained in the functions
Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) are regular on H. Therefore these functions
satisfy condition 1) and by Lemma 1 also condition 4) of the Definition
from [1] (p. 53).
II. From (1.6) it follows that
Γ0 (4N ) ⊂ Γ0 (4Nk ) (k = 1, 2, . . . , s − 8).
(1.9)
It is easy to verify that (1.7) implies
s−8
s−8 2
X
X
gk2 2ϕ(2Nk )−2
hk
, 4
δ
4N δ 2
,
Nk
4Nk
(1.10)
k=1
k=1
since 2 ∤ δ by virtue of αδ ≡ 1 (mod 4N ).
Using (1.9), (1.10) and Lemma 3 from [1] (p. 58), for all substitutions
from Γ0 (4N ) we get:
1) if n = 3, n = 1 and n = 0,
ατ + β
ατ + β
; 0, 2Nr ϑ′gt ht
; 0, 2Nt ×
ϑ′′′
gr hr
γτ + δ
γτ + δ
s−8
Y
ατ + β
ϑgk hk
; 0, 2Nk =
×
γτ + δ
k=3
=i
sη(γ)(sgn δ−1)/2
(s−8)(1−|δ|)/2
·i
s−8
Q
Nk
(γτ + δ)s/2 ×
|δ|
k=1
×ϑ′′′
αgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt )
s−8
Y
k=3
for r = 1, t = 2 and r = 2, t = 1;
ϑαgk ,hk (τ ; 0, 2Nk )
(1.11)
474
G. LOMADZE
2) if n = 4 and n = 0,
(4)
ϑgr hr
ατ + β
ατ + β
; 0, 2Nr ϑgt ht
; 0, 2Nt ×
γτ + δ
γτ + δ
s−8
Y
ατ + β
; 0, 2Nk =
×
ϑgk hk
γτ + δ
k=3
=i
sη(γ)(sgn δ−1)/2
(s−8)(1−|δ|)/2
·i
s−8
Q
Nk
(γτ + δ)s/2 ×
|δ|
k=1
(4)
×ϑαgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt )
s−8
Y
ϑαgk ,hk (τ ; 0, 2Nk )
(1.12)
k=3
for r = 1, t = 2 and r = 2, t = 1;
3) if n = 2 and n = 0,
ϑg′′1 h1
ατ + β
ατ + β
; 0, 2N1 ϑ′′g2 h2
; 0, 2N2 ×
γτ + δ
γτ + δ
s−8
ατ + β
Y
; 0, 2Nk =
ϑgk hk
×
γτ + δ
k=3
= isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2
s−8
Q
Nk
(γτ + δ)s/2 ×
|δ|
k=1
×ϑ′′αg1 ,h1 (τ ; 0, 2N1 )ϑ′′αg2 ,h2 (τ ; 0, 2N2 )
s−8
Y
ϑαgk ,hk (τ ; 0, 2Nk ).
(1.13)
k=3
Thus by (1.1), (1.11) and (1.2), (1.12), (1.13) we have
Ψj
ατ + β
; gl , hl , 0, Nl = isη(γ)(sgn δ−1)/2 · i(s−8)(1−|δ|)/2 ×
γτ + δ
s−8
Q
Nk
× k=1
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) (j = 1, 2),
|δ|
which by Lemma 2 implies respectively for j = 1 and j = 2
Ψj
ατ + β
; gl , hl , 0, Nl =
γτ + δ
s−8
Q
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl )
|δ|
if s ≡ 0 (mod 4)
k=1
ON SOME ENTIRE MODULAR FORMS
475
and
Ψj
= sgn δ
−1
|δ|
ατ + β
; gl , hl , 0, Nl =
γτ + δ
s−8
Q
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 2 (mod 4).
|δ|
k=1
Hence by (1.5) and (1.8) we obtain
ατ + β
Ψj
; gl , hl , 0, Nl = χ(δ)(γτ + δ)s/2 Ψj (τ ; gl , hl , 0, Nl ).
γτ + δ
Thus the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) satisfy condition (2) of the
Definition from [1].
III. The functions Ψj (τ ; gl , hl , 0, Nl ) (j = 1, 2) also satisfy condition 3)
of the Definition from [1]. For the proof see [2] (p. 109).
2.
To simplify the notations of this section we put
Ω(τ ; g1 , h1 , c1 , N1 ; g2 , h2 , h2 , c2 , N2 ) =
1 ′′
=
(τ ; c1 , 2N1 )ϑg2 h2 (τ ; c2 , 2N2 ) −
ϑ
N1 g1 h1
1 ′′
−
ϑ
(τ ; c2 , 2N2 )ϑg1 h1 (τ ; c1 , 2N1 )
N2 g2 h2
(2.1)
and occasionally
(s − 2)/3 = se, (s − 1)/3 = s′ .
(2.2)
Lemma 3. For even s ≥ 10 and given N let
Ψ3 (τ ; gl , hl , cl , Nl ) = Ω(τ ; g1 , h1 , c1 , N1 ; g2 , h2 , c2 , N2 ) ×
×
s/3
Y
ϑ′gk hk (τ ; ck , 2Nk ) if s ≡ 0 (mod 6),
(2.3)
k=3
Ψ3 (τ ; gl , hl , cl , Nl ) = Ω(τ ; g1 , h1 , c1 , N1 ; g2 , h2 , c2 , N2 ) ×
×
s˜
Y
ϑg′ k hk (τ ; ck , 2Nk )
k=3
+2
s˜
Y
ϑgk hk (τ ; ck , 2Nk ) if s ≡ 2 (mod 6),
(2.4)
s+1
k=˜
Ψ3 (τ ; gl , hl , cl , Nl ) = Ω(τ ; g1 , h1 , c1 , N1 ; g2 , h2 , c2 , N2 ) ×
′
×
s
Y
k=3
ϑ′gk hk (τ ; ck , 2Nk )ϑgs′ +1 hs′ +1 (τ ; cs′ +1 , 2Ns′ +1 ) if s ≡ 4 (mod 6) (2.5)
476
G. LOMADZE
and
Ψ4 (τ ; gl , hl , cl , Nl ) =
s/3
Y
ϑ′gk hk (τ ; ck , 2Nk ) if s ≡ 0 (mod 6),
(2.6)
k=1
Ψ4 (τ ; gl , hl , cl , Nl ) =
s˜
Y
ϑ′gk hk (τ ; ck , 2Nk )ϑgs˜+1 hs˜+1 (τ ; cs˜+1 , 2Ns˜+1 ) ×
k=1
×ϑgs˜+2 hs˜+2 (τ ; cs˜+2 , 2Ns˜+2 ) if s ≡ 2 (mod 6),
(2.7)
′
Ψ4 (τ ; gl , hl , cl , Nl ) =
s
Y
ϑ′gk hk (τ ; ck , 2Nk )ϑgs′ +1 hs′ +1 (τ ; cs′ +1 , 2Ns′ +1 ) (2.8)
k=3
if s ≡ 4 (mod 6),
where
s/3
X hk
4N
if s ≡ 0 (mod 6),
Nk
2|gk , Nk |N (k = 1, 2, . . . , s/3),
(2.9)
k=1
2|gk , Nk |N (k = 1, 2, . . . , se + 2),
2|gk , Nk |N (k = 1, 2, . . . , s′ + 1),
s˜+2
X
hk
if s ≡ 2 (mod 6), (2.10)
4N
Nk
k=1
′
+1
sX
hk
if s ≡ 4 (mod 6), (2.11)
4N
Nk
k=1
Then we have
(γτ + δ)s/2 Ψj (τ ; gl , hl , 0, Nl ) =
n ατ + β
(j = 3, 4)
Cn(j) e
4N γτ + δ
n=0
∞
X
(2.12)
for all substitutions from Γ in the neighborhood of each point τ = − γδ (γ 6= 0,
(γ, δ) = 1).
Proof. I. It is shown in [3] (formula (1.7)) that
(γτ + δ)3 Ω(τ ; g1 , h1 , 0, N1 ; g2 , h2 , 0, N2 ) =
3
= −e
sgn γ (4N1 N2 γ 2 )−1/2
4
×Ω
X
2
Y
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
H1 mod 2N1 k=1
H2 mod 2N2
ατ + β
; g1′ h1′ , H1 , N1 ; g2′ , h′2 , H2 , 2N2 .
γτ + δ
For n = 1 and n = 0 it follows by Lemma 4 from [1] (p. 61) that:
(2.13)
ON SOME ENTIRE MODULAR FORMS
(γτ + δ)(s−6)/2
1)
s/3
Y
477
ϑg′ k hk (τ ; 0, 2Nk ) =
k=3
× 2(s−6)/3
= e (s − 6)(sgn γ)/8 (−i sgn γ)(s−6)/3 ×
s/3
s/3
−1/2
Y
Y
X
(s−6)/3
Nk |γ|
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
k=3
H3 mod 2N3 k=3
........................
Hs/3 mod 2Ns/3
ατ + β
if s ≡ 0 (mod 6),
; Hk , 2Nk
×ϑgk′ h′k
γτ + δ
2)
(γτ + δ)(s−6)/2
s˜
Y
(2.14)
ϑ′gk hk (τ ; 0, 2Nk )ϑgs˜+1 hs˜+1 (τ ; 0, 2Ns˜+1 )×
k=3
×ϑgs˜+2 hs˜+2 (τ ; 0, 2Ns˜+2 ) = e (s − 6)(sgn γ)/8 ×
s˜+2
−1/2
X
× 2s˜
(−i sgn γ)(s−8)/3 ×
Nk |γ|s˜
k=3
s˜
Y
X
×
k
k=3
H3 mod 2N3
........................
Hs˜+2 mod 2Ns˜+2
×
s˜
+2
Y
k=˜
s+1
ατ + β
×
; Hk , 2Nk
k
γτ + δ
ϕgk′ gk hk (0, Hk ; 2Nk )ϑ′g′ h′
ϕgk′ gk hk (0, Hk ; 2Nk )ϑgk′ h′k
ατ + β
; Hk , 2Nk
γτ + δ
(2.15)
if s ≡ 2 (mod 6),
′
3)
(s−6)/2
(γτ + δ)
s
Y
ϑ′gk hk (τ ; 0, 2Nk )ϑgs′ +1 hs′ +1 (τ ; 0, 2Ns′ +1 ) =
k=3
+1
−1/2
′ sX
′
= e (s − 6)(sgn γ)/8 2s −1
Nk |γ|s −1
(−i sgn γ)(s−7)/3 ×
′
k=3
×
X
s′
Y
k=3
H3 mod 2N3
...........................
Hs′ +1 mod 2Ns′ +1
ατ + β
; Hk , 2Nk ×
k
γτ + δ
ϕgk′ gk hk (0, Hk ; 2Nk )ϑg′ ′ h′
k
×ϕg′ ′ gs′ +1 hs′ +1 (0, Hs′ +1 ; 2Ns′ +1 ) ×
s +1
ατ + β
×ϑgs′ +1 hs′ +1
; Hs′ +1 , 2Ns′ +1 if s ≡ 4 (mod 6).
γτ + δ
(2.16)
478
G. LOMADZE
After multiplying (2.13) successively by (2.14), (2.15) and (2.16), by (2.3)
we obtain
1) (γτ + δ)s/2 Ψ3 (τ ; gl , hl , 0, Nl ) = (−1)s/6 e s(sgn γ)/8 ×
s/3
−1/2
Y
s/3
Nk |γ|s/3
× 2
k=1
s/3
Y
X
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
H1 mod 2N1 k=3
........................
Hs/3 mod 2Ns/3
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 0 (mod 6),
γτ + δ
(γτ + δ)s/2 Ψ3 (τ ; gl , hl , 0, Nl ) = (−1)(s−2)/6 e s(sgn γ)/8 ×
×Ψ3
2)
s˜
+2
−1/2
Y
Nk |γ|s˜+2
× 2s˜+2
k=1
s˜
+2
Y
X
(2.17)
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
k=3
H1 mod 2N1
........................
Hs˜+2 mod 2Ns˜+2
ατ + β
(2.18)
; gl′ , h′l , Hl , Nl if s ≡ 2 (mod 6),
γτ + δ
(γτ + δ)s/2 Ψ3 (τ ; gl , hl , 0, Nl ) = (−1)(s+2)/6 (i sgn γ)e s(sgn γ)/8 ×
×Ψ3
3)
s′ +1
× 2
′
+1
sY
Nk |γ|
s′ +1
k=1
×Ψ3
−1/2
′
sY
+1
X
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
H1 mod 2N1
k=3
...........................
Hs′ +1 mod 2Ns′ +1
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 4 (mod 6).
γτ + δ
(2.19)
Further, reasoning as in [1] (Lemma 5, the bottom part of p. 64 and p.
65), we obtain (2.12) if j = 3.
II. By Lemma 4 from [1], for n = 1 and n = 0 it respectively follows by
(2.6), (2.7) and (2.8) that
s/2
(γτ + δ)
s/2
Ψ4 (τ ; gl , hl , 0, Nl ) = (γτ + δ)
s/3
Y
ϑ′gk hk (τ ; 0, 2Nk ) =
k=1
= e s(sgn γ)/8
×
X
s/3
2s/3
Y
k=1
Nk |γ|s/3
−1/2
(−i sgn γ)s/3 ×
ατ + β
Y
′
=
ϕgk′ gk hk (0, Hk ; 2Nk )ϑg′ h′
; Hk , 2Nk
k k
γτ + δ
s/3
H1 mod 2N1 k=1
........................
Hs/3 mod 2Ns/3
ON SOME ENTIRE MODULAR FORMS
479
s/3
−1/2
Y
×
= (−1)s/6 e s(sgn γ)/8 2s/3
Nk |γ|s/3
k=1
X
×
s/3
Y
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
H1 mod 2N1 k=1
........................
Hs/3 mod 2Ns/3
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 0 (mod 6),
(2.20)
γτ + δ
s˜
Y
(γτ + δ)s/2 Ψ4 (τ ; gl , hl , 0, Nl ) = (γτ + δ)s/2
ϑg′ k hk (τ ; 0, 2Nk ) ×
×Ψ4
k=1
×ϑgs˜+1 hs˜+1 (τ ; 0, 2Ns˜+1 )ϑgs˜+2 hs˜+2 (τ ; 0, 2Ns˜+2 ) =
s˜
+2
−1/2
Y
(−i sgn γ)s˜ ×
Nk |γ|s˜+2
= e s(sgn γ)/8 2s˜+2
k=1
×
s˜
Y
X
H1 mod 2N1 k=1
..................
Hs˜ mod 2Ns˜
×
ατ + β
×
; Hk , 2Nk
ϕgk′ gk hk (0, Hk ; 2Nk )ϑ′g′ h′
k k
γτ + δ
+2
s˜
Y
X
s+1
Hs˜+1 mod 2Ns˜+1 k=˜
Hs˜+2 mod 2Ns˜+2
ατ + β
=
; Hk , 2Nk
ϕgk′ gk hk (0, Hk ; 2Nk )ϑgk hk
γτ + δ
s˜
+2
−1/2
Y
= (−1)(s−2)/6 e s(sgn γ)/8 2s˜+2
×
Nk |γ|s˜+2
k=1
×
X
s˜
+2
Y
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
k=1
H1 mod 2N1
........................
Hs˜+2 mod 2Ns˜+2
×Ψ4
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 2 (mod 6)
γτ + δ
(2.21)
and
′
(γτ + δ)
s/2
Ψ4 (τ ; gl , hl , 0, Nl ) = (γτ + δ)
s/2
s
Y
ϑg′ k hk (τ ; 0, 2Nk ) ×
k=1
×ϑgs′ +1 hs′ +1 (τ ; 0, 2Ns′ +1 )e s(sgn γ)/8 ×
′
+1
−1/2
′ sY
′
′
s +1
(−i sgn γ)s ×
Nk |γ|s +1
× 2
k=1
480
×
G. LOMADZE
s′
ατ + β
Y
′
×
; Hk , 2Nk
ϕgk′ gk hk (0, Hk ; 2Nk )ϑg′ h′
k k
γτ + δ
X
H1 mod 2N1 k=1
.....................
Hs′ mod 2Ns′
X
×
′
ϕgs+1
gs′ +1 hs′ +1 (0, Hs′ +1 ; 2Ns′ +1 ) ×
Hs′ +1 mod 2Ns′ +1
ατ + β
; Hs′ +1 , 2Ns′ +1 =
γτ + δ
′
+1
−1/2
s′ +1 sY
′
(s+2)/6
(i sgn γ)e s(sgn γ)/8 2
= (−1)
Nk |γ|s +1
×
×ϑgs′ +1 hs′ +1
k=1
×
X
′
sY
+1
ϕgk′ gk hk (0, Hk ; 2Nk ) ×
k=1
H1 mod 2N1
.....................
Hs′ +1 mod 2Ns′ +1
×Ψ4
ατ + β
; gl′ , h′l , Hl , Nl if s ≡ 4 (mod 6).
γτ + δ
(2.22)
Further, reasoning as in [1] (Lemma 5), we obtain (2.12) if j = 4.
Lemma 4. For even s we have
(a) isη(γ)(sgn δ−1)/2 · i(s/3)(1−|δ|)/2 =
1
−1
if s ≡ 0 (mod 12),
sgn δ
if s ≡ 6 (mod 12);
|δ|
1
if s ≡ 8 (mod 12),
−1
(b) isη(γ)(sgn δ−1)/2 · i((s+4)/3)(1−|δ|)/2=
sgn δ
if s ≡ 2 (mod 12);
|δ|
−1
if s ≡ 4 (mod 12),
sη(γ)(sgn δ−1)/2
((s+2)/3)(1−|δ|)/2
|δ|
·i
(c) i
=
sgn δ
if s ≡ 10 (mod 12).
Proof. (a) Let s ≡ 0 (mod 6), i.e., s ≡ 0 (mod 12) or s ≡ 6 (mod 12).
Then
isη(γ)(sgn δ−1)/2 · i(s/3)(1−|δ|)/2 = i3hη(γ)(sgn δ−1) · i2h(1−|δ|)/2 =
1
if 2|h, i.e., if s ≡ 0 (mod 12),
−1
=
sgn δ
if 2 ∤ h, i.e., if s ≡ 6 (mod 12).
|δ|
ON SOME ENTIRE MODULAR FORMS
481
(b) Let s ≡ 2 (mod 6), i.e., s ≡ 2 (mod 12) or s ≡ 8 (mod 12). Then
isη(γ)(sgn δ−1)/2 · i((s+4)/3)(1−|δ|)/2 = i(3h+1)η(γ)(sgn δ−1) · i2(h+1)(1−|δ|)/2 =
= i(h+1)η(γ)(sgn δ−1) · (−1)(h+1)(1−|δ|)/2 =
sgn δ −1
if 2|h, i.e., if s ≡ 2 (mod 12),
|δ|
=
1
if 2 ∤ h, i.e., if s ≡ 8 (mod 12).
(c) Let s ≡ 4 (mod 6), i.e., s ≡ 4 (mod 12) or s ≡ 10 (mod 12). Then
isη(γ)(sgn δ−1)/2 · i((s+2)/3)(1−|δ|)/2 = i(3h+2)η(γ)(sgn δ−1) · i2(h+1)(1−|δ|)/2 =
−1
if 2|h, i.e., if s ≡ 4 (mod 12),
|δ|
=
if 2 ∤ h, i.e., if s ≡ 10 (mod 12).
sgn δ
Theorem 2. For even s and given N the functions Ψj (τ ; gl , hl , 0, Nl )
(j = 3, 4) are entire modular forms of weight 2s and the character
∆
if s ≡ 0 (mod 4),
|δ|
−∆
(2.23)
χ(δ) =
sgn δ
if s ≡ 2 (mod 4)
|δ|
(∆ is the determinant of the positive quadratic form with integral coefficients
in s variables for which the function Ψj will be used) for the congruence
subgroup Γ0 (4N ) if the following conditions hold:
(a) when s ≡ 0 (mod 6)
1) 2|gk , Nk |N (k = 1, 2, . . . , s/3),
s/3
X
hk2
,
2) 4N
Nk
k=1
(2.24)
s/3
X
gk2
4
,
4Nk
(2.25)
k=1
3) for all α and δ with αδ ≡ 1 (mod 4N )
s/3
Q
k=1
Nk
|δ|
Ψj (τ ; αgl , hl , 0, Nl ) =
∆
|δ|
Ψj (τ ; gl , hl , 0, Nl );
(2.26)
(b) when s ≡ 2 (mod 6)
1) 2|gk , Nk |N (k = 1, 2, . . . , (s + 4)/3),
(s+4)/3
(s+4)/3
X h2
X gk2
k
, 4
,
2) 4N
Nk
4Nk
k=1
k=1
(2.27)
(2.28)
482
G. LOMADZE
3) for all α and δ with αδ ≡ 1 (mod 4N )
(s+4)/3
Q
Nk
∆
Ψj (τ ; αgl , hl , 0, Nl ) =
Ψj (τ ; gl , hl , 0, Nl );
|δ|
|δ|
k=1
(2.29)
(c) when s ≡ 4 (mod 6)
1) 2|gk , Nk |N (k = 1, 2, . . . , (s + 2)/3),
(s+2)/3
X h2
k
2) 4N
,
Nk
k=1
(s+2)/3
X gk2
4
,
4Nk
(2.30)
(2.31)
k=1
3) for all α and δ with αδ ≡ 1 (mod 4N )
(s+2)/3
Q
k=1
|δ|
Nk
Ψj (τ ; αgl , hl , 0, Nl ) = sgn δ
−∆
Ψj (τ ; gl , hl , 0, Nl ). (2.32)
|δ|
Proof. I. As in the case of Theorem 1, the functions Ψj (τ ; gl , hl , 0, Nl ) (j =
3, 4) satisfy condition 1) and by Lemma 3 also condition 4) of the Definition
from [1].
II. From (2.24), (2.27) and (2.30) we obtain
Γ0 (4N ) ⊂ Γ0 (4Nk ).
(2.33)
Obviously, (2.25), (2.28) and (2.31) respectively imply
s/3
2 X h2k
,
4N δ
Nk
k=1
(s+4)/3 2
X h
k
4 N δ 2
,
Nk
k=1
and
(s+2)/3 2
X h
k
,
4N δ 2
Nk
k=1
s/3
X
gk2 2ϕ(2Nk )−2
δ
if s ≡ 0 (mod 6),
4
4Nk
(2.34)
k=1
(s+4)/3
X gk2 2ϕ(2Nk )−2
δ
4
if s ≡ 2 (mod 6), (2.35)
4Nk
k=1
(s+2)/3
X gk2 2ϕ(2Nk )−2
δ
4
if s ≡ 4 (mod 6).
4Nk
(2.36)
k=1
Using (2.33) and Lemma 3 from [1], for all substitutions from Γ0 (4N ) we
obtain:
1) for n = 2, n = 1 and n = 0
ατ + β
ατ + β
ϑ′′gr hr
; 0, 2Nr ϑgt ht
; 0, 2Nt ×
γτ + δ
γτ + δ
ON SOME ENTIRE MODULAR FORMS
×
s/3
Y
k=3
ϑ′gk hk
483
ατ + β
; 0, 2Nk =
γτ + δ
s/3
Q
Nk
(γτ + δ)s/2 ×
·i
|δ|
×ϑ′′αgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt ) ×
sη(γ)(sgn δ−1)/2
=i
×
s/3
Y
(s/3)(1−|δ|)/2
k=1
′
(τ ; 0, 2Nk ) if s ≡ 0 (mod 6),
ϑαg
k ,hk
(2.37)
k=3
ατ + β
ατ + β
; 0, 2Nr ϑgt ht
; 0, 2Nt ×
γτ + δ
γτ + δ
s
˜
ατ + β
Y
ϑ′gk hk
×
; 0, 2Nk ×
γτ + δ
k=3
ατ + β
ατ + β
; 0, 2Ns˜+1 ϑgs˜+2 hs˜+2
; 0, 2Ns˜+2 =
×ϑgs˜+1 hs˜+1
γτ + δ
γτ + δ
s˜Q
+2
Nk
= isη(γ)(sgn δ−1)/2 · i(˜s+2)(1−|δ|)/2 k=1
(γτ + δ)s/2 ×
|δ|
ϑ′′gr hr
′′
×ϑαg
(τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt )
r ,hr
s˜
Y
ϑ′αgk ,hk (τ ; 0, 2Nk ) ×
k=3
×ϑαgs˜+1 ,hs˜+1 (τ ; 0, 2Ns˜+1 )ϑαgs˜+2 ,hs˜+2 (τ ; 0, 2Ns˜+2 )
(2.38)
if s ≡ 2 (mod 6),
ατ + β
ατ + β
ϑg′′r hr
; 0, 2Nr ϑgt ht
; 0, 2Nt ×
γτ + δ
γτ + δ
′
s
ατ + β
Y
ατ + β
; 0, 2Nk ϑgs′ +1 hs′ +1
; 0, 2Ns′ +1 =
×
ϑg′ k hk
γτ + δ
γτ + δ
k=3
′
= (sgn δ)isη(γ)(sgn δ−1)/2 · i(s +1)(1−|δ|)/2 ×
′
s
Q
Nk
k=1
×
(γτ + δ)s/2 ϑ′′αgr ,hr (τ ; 0, 2Nr )ϑαgt ,ht (τ ; 0, 2Nt ) ×
|δ|
′
×
s
Y
ϑαgk ,hk (τ ; 0, 2Nk )ϑαgs′ +1 ,hs′ +1 (τ ; 0, 2Ns′ +1 )
k=3
if s ≡ 4 (mod 6),
(2.39)
484
G. LOMADZE
where r = 2, t = 1 and r = 1, t = 2;
2) for n = 1 and n = 0
s/3
Y
ϑ′gk hk
k=1
ατ + β
, 0, 2Nk =
γτ + δ
= isη(γ)(sgn δ−1)/2 · i(s/3)(1−|δ|)/2
s/2
×(γτ + δ)
s/3
Y
s/3
Q
k=1
Nk
|δ|
×
ϑ′αgk ,hk (τ ; 0, 2Nk ) if s ≡ 0 (mod 6),
(2.40)
k=1
s˜
Y
ϑ′gk hk
k=1
ατ + β
ατ + β
, 0, 2Nk ϑgs˜+1 hs˜+1
, 0, 2Ns˜+1 ×
γτ + δ
γτ + δ
ατ + β
, 0, 2Ns˜+2 = isη(γ)(sgn δ−1)/2 · i(˜s+2)(1−|δ|)/2 ×
γτ + δ
s˜Q
+2
Nk
s˜
Y
k=1
×
ϑ′αgk ,hk (τ ; 0, 2Nk ) ×
(γτ + δ)s/2
|δ|
×ϑgs˜+2 hs˜+2
k=1
×ϑαgs˜+1 ,hs˜+1 (τ ; 0, 2Ns˜+1 )ϑαgs˜+2 ,hs˜+2 (τ ; 0, 2Ns˜+2 )
(2.41)
if s ≡ 2 (mod 6),
s′
Y
ϑ′gk hk
k=1
ατ + β
ατ + β
, 0, 2Nk ϑgs′ +1 hs′ +1
, 0, 2Ns′ +1 =
γτ + δ
γτ + δ
′
= sgn δisη(γ)(sgn δ−1)/2 · i(s +1)(1−|δ|)/2
′
sQ
+1
k=1
Nk
|δ|
(γτ + δ)s/2 ×
′
×
s
Y
ϑ′αgk hk (τ ; 0, 2Nk )ϑαgs′ +1 ,hs′ +1 (τ ; 0, 2Ns′ +1 )
k=1
if s ≡ 4 (mod 6).
By (2.3), (2.37) and (2.6), (2.40), for j = 3 and j = 4 we get
Ψj
ατ + β
; gl , hl , 0, Nl = isη(γ)(sgn δ−1)/2 · i(s/3)(1−|δ|)/2 ×
γτ + δ
(2.42)
ON SOME ENTIRE MODULAR FORMS
×
485
s/3
Q
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 0 (mod 6). (2.43)
|δ|
k=1
By (2.4), (2.38) and (2.7), (2.41), for j = 3 and j = 4 we get
Ψj
×
ατ + β
; gl , hl , 0, Nl = isη(γ)(sgn δ−1)/2 · i((s+4)/3)(1−|δ|)/2 ×
γτ + δ
(s+4)/3
Q
k=1
|δ|
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 2 (mod 6). (2.44)
By (2.5), (2.39) and (2.8), (2.42), for j = 3 and j = 4 we get
Ψj
×
ατ + β
; gl , hl , 0, Nl = isη(γ)(sgn δ−1)/2 · i((s+2)/3)(1−|δ|)/2 ×
γτ + δ
(s+2)/3
Q
k=1
|δ|
Nk
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 4 (mod 6). (2.45)
Further, by Lemma 4, for j = 3, 4 realtions (2.43), (2.44), and (2.45)
respectively imply:
s/3
Q
Nk
ατ + β
k=1
Ψj
(γτ + δ)s/2 Ψj (τ ; αgl , hl , 0, Nl )
; gl , hl , 0, Nl =
γτ + δ
|δ|
if s ≡ 0 (mod 12),
−1
s/3
Q
k=1
(2.46)
Nk
(γτ + δ)s/2 ×
|δ|
|δ|
× Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 6 (mod 12);
= sgn δ
Ψj
(s+4)/3
Q
Nk
−1
ατ + β
k=1
(γτ + δ)s/2 ×
; gl , hl , 0, Nl = sgn δ
γτ + δ
|δ|
|δ|
× Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 2 (mod 12),
=
(s+4)/3
Q
k=1
|δ|
(2.47)
Nk
s/2
(γτ + δ) Ψj (τ ; αgl , hl , 0, Nl )
if s ≡ 8 (mod 12);
486
G. LOMADZE
(s+2)/3
Q
Nk
ατ + β
−1
k=1
(γτ + δ)s/2 ×
; gl , hl , 0, Nl = sgn δ
Ψj
γτ + δ
|δ|
|δ|
× Ψj (τ ; αgl , hl , 0, Nl ) if s ≡ 4 (mod 12),
=
(s+2)/3
Q
k=1
|δ|
(2.48)
Nk
s/2
(γτ + δ) Ψj (τ ; αgl , hl , 0, Nl )
if s ≡ 10 (mod 12).
From (2.46)–(2.48), (2.26), (2.29), (2.32), and (2.23) it follows that
Ψj
ατ + β
; gl , hl , 0, Nl = χ(δ)(γτ + δ)s/2 Ψj (τ ; gl , hl , 0, Nl ) (j = 3, 4).
γτ + δ
Thus the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 3, 4) satisfy condition 2) of the
Definition from [1].
III. By [1] (formula 1.3), for r = 1, t = 2 and r = 2, t = 1 we obtain:
1) If s ≡ 0 (mod 6),
ϑ′′gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )
s/3
Y
ϑ′gk hk (τ ; 0, 2Nk ) =
k=3
= (πi)s/3
∞
X
hr mr +ht mt +
(−1)
s/3
P
k=3
hk mk
(4Nr mr + gr )2 ×
mr ,mt ,m3 ,...,ms/3 =−∞
×
s/3
Y
k=3
X
X
s/3
∞
gk 2
1
=
2Nk mk +
Bn(r,t) e(nτ ),
(4Nk mk + gk )e
4Nk
2
n=0
k=1
since by (2.24) and (2.25),
n=
s/3
s/3
s/3
X
1X 2
1
gk 2 X
Nk m2k + mk gk /2 +
gk /4Nk
=
2Nk mk +
4Nk
2
4
k=1
k=1
k=1
is a non-negative integer;
2) If s ≡ 2 (mod 6),
ϑg′′r hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt )
s˜
Y
ϑ′gk hk (τ ; 0, 2Nk ) ×
k=3
×ϑgs˜+1 hs˜+1 (τ ; 0, 2Ns˜+1 )ϑgs˜+2 hs˜+2 (τ ; 0, 2Ns˜+2 ) =
ON SOME ENTIRE MODULAR FORMS
P
487
˜+2
s
∞
X
s˜
= (πi)
hr mr +ht mt +
(−1)
hk mk
k=3
(4Nr mr + gr )2 ×
mr ,mt ,m3 ,...,ms˜+2 =−∞
s˜
Y
×
k=3
X
X
s˜+2
∞
1
gk 2
(4Nk mk + gk )e
=
Cn(r,t) e(nτ ),
2Nk mk +
4Nk
2
n=0
k=1
since by (2.27) and (2.28),
n=
s˜+2
s˜+2
s˜+2
X
1X
gk 2 X
1
gk2 /4Nk
2Nk mk +
Nk m2k + mk gk /2 +
=
4Nk
2
4
k=1
k=1
k=1
is a non-negative integer;
3) If s ≡ 4 (mod 6),
ϑ′′gr hr (τ ; 0, 2Nr )ϑgt ht (τ ; 0, 2Nt ) ×
′
×
s
Y
ϑ′gk hk (τ ; 0, 2Nk )ϑgs′ +1 hs′ +1 (τ ; 0, 2Ns′ +1 ) ×
k=3
P
s′ +1
= (πi)
∞
X
′
s
hr mr +ht mt +
(−1)
k=3
h k mk
(4Nr mr + gr )2 ×
mr ,mt ,m3 ,...,ms′ +1 =−∞
′
×
s
Y
k=3
′
X
sX
∞
+1
gk 2
1
Dn(r,t) e(nτ ),
=
2Nk mk +
(4Nk mk + gk )e
4Nk
2
n=0
k=1
since By (2.30) and (2.31)
n=
′
sX
+1
k=1
′
+2
s′ +1
1 sX
gk 2 X
1
2
2Nk mk +
gk2 /4Nk
=
Nk mk + mk gk /2 +
4Nk
2
4
k=1
k=1
is a non-negative integer.
Analogously,
s/3
Y
ϑg′ k hk (τ ; 0, 2Nk ) =
k=1
s/3
= (πi)
∞
X
m1 ,m2 ,...,ms/3 =−∞
(−1)
s/3
P
k=1
s/3
hk mk Y
(4Nk mk + gk ) ×
k=1
s/3
∞
X
2 X
1
×e
Bn e(nτ ) if s ≡ 0 (mod 6),
2Nk mk + gk /2 τ =
4Nk
n=0
k=1
488
G. LOMADZE
s˜
Y
ϑ′gk hk (τ ; 0, 2Nk )ϑgs˜+1 hs˜+1 (τ ; 0, 2Ns˜+1 )ϑgs˜+2 hs˜+2 (τ ; 0, 2Ns˜+2 ) =
k=1
P
˜+2
s
∞
X
= (πi)s˜
hk mk
(−1)k=1
m1 ,m2 ,...,ms˜+2 =−∞
s˜
Y
(4Nk mk + gk ) ×
k=1
s˜+2
∞
X
2 X
1
×e
2Nk mk + gk /2 τ =
Cn e(nτ ) if s ≡ 2 (mod 6),
4Nk
n=0
k=1
′
s
Y
ϑg′ k hk (τ ; 0, 2Nk )ϑgs′ +1 hs′ +1 (τ ; 0, 2Ns′ +1 ) =
k=1
P
s′ +1
s
= (πi)
′
∞
X
(−1)
k=1
m1 ,m2 ,...,ms′ +1 =−∞
s′
hk mk Y
(4Nk mk + gk ) ×
k=1
′
∞
+1
sX
2 X
1
2Nk mk + gk /2 τ =
Dn e(nτ ) if s ≡ 4 (mod 6).
×e
4Nk
n=0
k=1
Thus by (2.3)–(2.8) the functions Ψj (τ ; gl , hl , 0, Nl ) (j = 3, 4) satisfy
condition 3) of the Definition from [1].
References
1. G. Lomadze, On some entire modular forms of weights 5 and 6 for the
congruence group Γ0 (4N ). Georgian Math. J. 1(1994), No. 1, 53–76.
2. G. Lomadze, On some entire modular forms of half-integral weight for
the congruence group Γ0 (4N ). Proc. A. Razmadze Math. Inst. 118(1998),
101–109.
3. G. Lomadze, Construction of entire modular forms of weights 5 and
6 for the congruence group Γ0 (4N ). Georgian Math. J. 2(1995), No. 2,
189–199.
(Received 14.04.1998)
Author’s address:
Faculty of Mechanics and Mathematics
I. Javakhishvili Tbilisi State University
2, University St., Tbilisi 380043
Georgia